Restricted Measure is Measure

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.


Then the restricted measure $\mu \restriction_{\Sigma'}$ is a measure on the measurable space $\left({X, \Sigma'}\right)$.


Proof

Verify the axioms for a measure in turn for $\mu \restriction_{\Sigma'}$:


Axiom $(1)$

The statement of axiom $(1)$ for $\mu \restriction_{\Sigma'}$ is:

$\forall E' \in \Sigma': \mu \restriction_{\Sigma'} \left({E'}\right) \ge 0$


Now, for every $E' \in \Sigma'$, compute:

\(\displaystyle \mu \restriction_{\Sigma'} \left({E'}\right)\) \(=\) \(\displaystyle \mu \left({E'}\right)\) Definition of $\mu \restriction_{\Sigma'}$
\(\displaystyle \) \(\ge\) \(\displaystyle 0\) $\mu$ is a measure

$\Box$


Axiom $(2)$

Let $\left({E'_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma'$.

Then the statement of axiom $(2)$ for $\mu \restriction_{\Sigma'}$ is:

$\displaystyle \mu \restriction_{\Sigma'} \left({\bigcup_{n \mathop \in \N} E'_n}\right) = \sum_{n \mathop \in \N} \mu \restriction_{\Sigma'} \left({E'_n}\right)$


One can show this by means of the following computation:

\(\displaystyle \mu \restriction_{\Sigma'} \left({\bigcup_{n \mathop \in \N} E'_n}\right)\) \(=\) \(\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E'_n}\right)\) Definition of $\mu \restriction_{\Sigma'}$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \mu \left({E'_n}\right)\) $\mu$ is a measure
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \mu \restriction_{\Sigma'} \left({E'_n}\right)\) Definition of $\mu \restriction_{\Sigma'}$

$\Box$


Axiom $(3')$

The statement of axiom $(3')$ for $\mu \restriction_{\Sigma'}$ is:

$\mu \restriction_{\Sigma'} \left({\varnothing}\right) = 0$


By Sigma-Algebra Contains Empty Set, $\varnothing \in \Sigma'$. Hence:

$\mu \restriction_{\Sigma'} \left({\varnothing}\right) = \mu \left({\varnothing}\right) = 0$

because $\mu$ is a measure.

$\Box$


Having verified a suitable set of axioms, it follows that $\mu \restriction_{\Sigma'}$ is a measure.

$\blacksquare$


Sources